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3 years ago

Find the area of the composite figure, rounding to the nearest tenth.

Mathematics

1 answer:

Ugo [173]3 years ago

7 0

Answer:

85,1 ft^2

Step-by-step explanation:

Semicircle area : 4^2π / 2 = 8π = 25,1 ft^2

Triangle area : (8 x 15) / 2 = 60 ft^2

Total Area = 60 + 25,1 = 85,1 ft^2

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Convert the following percent to a decimal: 56 percent.

mihalych1998 [28]

Answer:

0.56

Step-by-step explanation:

Percent means 'per 100'. So 56% means 56 per 100 or simply 56/100. If you divide 56 by 100, you'll get 0.56.

56/100 = 0.56

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2 years ago

The 12 students in the Environmental Club represent 10% of the students in the seventh grade. How many students are in the seven

Anna11 [10]

Answer:

Step-by-step explanation:

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2 years ago

What is the andwer???

djverab [1.8K]

Answer:

3 and 11

Step-by-step explanation:

Given 2 sides with third side x , then

difference of 2 sides < x < sum of 2 sides , that is

7 - 4 < x < 7 + 4

3 < x < 11

Thus the possible values the third side cannot be are 3 and 11

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3 years ago

Complete the following statement. 16 fl oz = pt

Elden [556K]

Answer:

Step-by-step explanation:

divide the volume value by 16

6 0

3 years ago

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

6 0

2 years ago